When those conditions are given in the diagram, What is the value of #sinx#? ABCD is a square.

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Answer 1 · 2017-10-08T11:29:16

# 3/5.#

We have, from the Right- #DeltaABP, AB^2+BP^2=AP^2.#

#:. AP^2=(2y)^2+y^2=5y^2.#

Similarly, #AQ^2=5y^2, and, PQ^2=2y^2.#

Applying the Cosine-Rule in #DeltaAPQ,# we get,

#cosx=(AP^2+AQ^2-PQ^2)/(2AP*AQ),#

#=(5y^2+5y^2-2y^2)/(2*sqrt5y*sqrt5y)=(8y^2)/(10y^2).#

# rArr cosx=4/5.#

Since #x# is acute, #sinx=+sqrt(1-cos^2x)=sqrt(1-16/25),# giving,

#sinx=3/5.#

Answer 2 · 2017-10-08T13:58:15

#sinx=3/5.#

Here is another Solution to the Problem :

Observe that #DeltaABP and DeltaADQ# are congruent.

#:. /_PAB=/_QAD........(1).#

From the Right- #DeltaABP, tan/_PAB=(PB)/(AP)=y/(2y).......(2).#

#:.," from "(1), and, (2), tan/_PAB=1/2=tan/_QAD.#

Since, #/_BAP+/_PAQ+/_QAD=pi/2,# we have.

#/_PAQ=pi/2-(/_BAP+/_QAD),#

#=pi/2-2/_BAP, i.e.,#

#x=pi/2-2theta, where, theta=/_BAP.#

#:. sinx=sin(pi/2-theta)=cos2theta=(1-tan^2theta)/(1+tan^2theta).#

But, #tantheta=tan/_BAP=1/2.#

# rArr sinx={1-(1/2)^2}/{1+(1/2)^}=(3/4)/(5/4)=3/5,# as Before!